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Moderator Note: This question is from a contest which ended 1 Dec 2012.

Consider a polynomial $f$ with complex coefficients. Call such $f$ broken if we can find a square matrix $M$ such that $M \neq f(N)$ for any square matrix $N$.

My professor told me it was possible to explicitly characterize all broken polynomials of any degree. What would such a characterization look like?

Thank you!

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Do you know how the eigenvalues of $f(N)$ relate to those of $N$? – Gerry Myerson Nov 7 '12 at 0:56
no, nothing more about $f$ is given than what was stated. How would the eigenvalues be of use though? – Andy Nov 7 '12 at 1:00
Ah I am having trouble finding that initial footing. Perhaps if you showed me how to start with that case, I may be able to proceed? – Andy Nov 7 '12 at 2:19
You have missed the intent of my question. Do you know any facts that, given any old matrix $A$, and any old polynomial $p$, relate the eigenvalues of $p(A)$ to those of $A$? – Gerry Myerson Nov 7 '12 at 6:02
Sorry, I am very new to this. My professor simply said this was a challenge problem, but I found it interesting. So if we apply $p$ to a matrix $A$, would that not change the eigenvalue? I apologize for not knowing too much about the subject – Andy Nov 7 '12 at 14:32

I don't have an answer, but I do have an example, and a few thoughts.

The problem is to find nonconstant $f$ in ${\bf C}[x]$ such that $f:M_n({\bf C})\to M_n({\bf C})$ is not onto.

Let's observe that for any nonconstant $f$ and any diagonal matrix $D$ there is a diagonal matrix $D'$ such that $f(D')=D$, simply because (as @Martin notes in the comments) every nonconstant polynomial is onto as a map from $\bf C$ to (shining) $\bf C$.

If $A$ is diagonalizable, say, $P^{-1}AP=D$ is diagonal, then find $D'$ such that $f(D')=D$, and you get $f(PD'P^{-1})=A$. This shows that if there's anything not in the range of $f$, it must be nondiagonalizable.

With that as motivating preamble, here's an example. I say $f(x)=x^2$ is "broken", because there's no matrix $N$ such that $N^2=M$, when $$M=\pmatrix{0&1\cr0&0\cr}$$

There's probably a conceptual proof of this, based on consideration of eigenspaces and such, but here's a brute force proof. If $$N=\pmatrix{a&b\cr c&d\cr}$$ then $$N^2=\pmatrix{a^2+bc&(a+d)b\cr(a+d)c&bc+d^2\cr}$$ So $N^2=M$ amounts to $$a^2+bc=0;\quad(a+d)b=1;\quad(a+d)c=0;\quad bc+d^2=0$$ The 3rd equation says $a+d=0$ or $c=0$; the second says $a+d\ne0$, hence, $c=0$. Then the first and fourth say $a=d=0$, and the second is impossible.

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I would be interested in seeing further advances on this question. I will try to get further with the question as well. – Xuan Huang Nov 22 '12 at 6:47

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